[method]

Hi all. i get the following error . could any one help me fixing it.Thanks

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in

error pointing to :

PHP Code:

                    $query = "INSERT INTO music (`ID`,`friendID`,`friendPicture`,`friendName`) VALUES ('$ID',' . $friend['id'] . ',' . $friend['picture'] . ',' . $friend['name'] . ')"; 

PHP Code:

....
....
foreach ($friends as $friend)
    {

        //here you can write to mysql


        if ($_POST['format'] == 'table')
                {
            fwrite($handle, '
            <tr>
                <td><a href="http://www.somsite.com/id=' . $friend['id'] . '">' . $friend['id'] . '</a></td>' . (!empty($_POST['photos']) ? '
                <td><img src="' . $friend['picture'] . '" alt="" /></td>' : '') . '
                <td>' . $friend['name'] . '</td>
            </tr>');


         

                    $query = "INSERT INTO music (`ID`,`friendID`,`friendPicture`,`friendName`) VALUES ('$ID',' . $friend['id'] . ',' . $friend['picture'] . ',' . $friend['name'] . ')";
                    $fp = fopen("videos.txt", "w");
                    fwrite($fp, $query);
                    fclose($fp);      
                    $result = mysql_query( $query );
                 }
        else
        
            fwrite($handle, '<li>
                <a href="http://www.somsite.com/id=' . $friend['id'] . '">
                    ' . (!empty($_POST['photos']) ? '<img src="' . $friend['picture'] . '" alt="" /><br />
                    ' : '') . $friend['name'] . '
                </a><br />
            </li>');
    } 

[wirehopper]

Change to:

PHP Code:

$query = "INSERT INTO music (`ID`,`friendID`,`friendPicture`,`friendName`) VALUES ('$ID','{$friend['id']}','{$friend['picture']}','{$friend['name']}')"; 

[method]

Thanks it worked well but only problem i get so many of the following notices

Notice: Undefined variable: ID in

and pointing to my insert statement. The data gets inserted to mysql but i get this strange notices. Could help me how to avoid them ?Thanks

[wirehopper]

ID is probably an autoincrement identifier, and thus shoud not be set.

Try removing it from the INSERT statement to see if it still works.

[method]

well i need the insert statement and if i remove it i will not get error !!
Is there away to keep the insert and not to get the error. it is just annyoing to get so many of such error

[method]

problem got solved by removing ID from insert statment :-)Thanks for all for helping me .

[Jay6390]

I'm fairly sure that the undefined variable notice had nothing to do with the database, it was because you had not initialised the $ID variable before the query line

[wirehopper]

The error message was issued because $ID was not initialized.

Many times, database ids are assigned by the database by setting the field up as an autoincrement field, in which case, assigning them is will also cause an error.

In order to be sure removing $ID from the INSERT command was the correct solution (and to avoid future problems) - you should check to see if the ID is an autoincrement field, and if it is supposed to be unique.